How do you find the derivative of $y = {[e^{- 1} + e^{t}]}^{3}$ $y = [e^(-1) + e^(t)]^3$ ?

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How do you find the derivative of $y = {[e^{- 1} + e^{t}]}^{3}$ $y = [e^(-1) + e^(t)]^3$ ?

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Answer 1 · 2015-05-15T18:27:42

The answer is : $3 {(e^{- 1} + e^{t})}^{2} \cdot e^{t}$ $3(e^(-1) + e^t)^2*e^t$

$. y = . {(e^{- 1} + e^{t})}^{3} = 3 {(e^{- 1} + e^{t})}^{2} \cdot . (e^{- 1} + e^{t})$ $dot y = dot [(e^(-1) + e^t)^3] = 3(e^(-1) + e^t)^2*dot ((e^(-1)+e^t))$

$= 3 {(e^{- 1} + e^{t})}^{2} \cdot e^{t}$ $= 3(e^(-1) + e^t)^2*e^t$

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How do you find the derivative of $y = {[e^{- 1} + e^{t}]}^{3}$ $y = [e^(-1) + e^(t)]^3$ ?

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How do you find the derivative of $y = {[e^{- 1} + e^{t}]}^{3}$ $y = [e^(-1) + e^(t)]^3$ ?